Question: Find $\tan\left(15^\circ\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$ (Choice B) B $\dfrac{\sqrt{3}-1}{2}$ (Choice C) C $\dfrac{\sqrt{3}-6}{\sqrt{3}+6}$ (Choice D) D $\dfrac{\sqrt{6}-\sqrt{2}}{4}$
Explanation: The strategy First, we should rewrite the given angle $15^\circ$ as the sum or difference of two special angles. Then, we can use the tangent addition or subtraction identities in order to evaluate $\tan\left(15^\circ\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $15^\circ$ We can rewrite $15^\circ$ as follows. $\begin{aligned}15^\circ&=60^\circ-45^\circ\end{aligned}$ In other words, $15^\circ$ is the difference of the special angles $60^\circ$ and $45^\circ$. Evaluating $\tan\left(15^\circ\right)$ Using the tangent subtraction identity, we get the following. $\begin{aligned} \tan\left(15^\circ\right)&= \tan\left(60^\circ-45^\circ\right) \\\\\\ &= \dfrac{\tan\left(60^\circ\right)-\tan\left(45^\circ\right)}{1+\tan\left(60^\circ\right)\tan\left(45^\circ\right)}\\\\\\ &= \dfrac{\left(\sqrt{3}\right)-\left(1\right)}{1+\left(\sqrt{3}\right)\left(1\right)}\\\\\\ &=\dfrac{\sqrt{3}-1}{\sqrt{3}+1} \end{aligned}$ Summary $\tan\left(15^\circ\right) = \dfrac{\sqrt{3}-1}{\sqrt{3}+1}$